Sometimes its possible to decompose the ST metric to two parts ## g=g_b+g_d ## where the difference between ## g_b ## and ## g_d ## lies in that you can define a time\length scale that ## g_d ## changes much faster than ## g_b ##. Because of this, you can call ## g_b ## the background metric and ## g_d ## a propagating distortion. Note that both parts of the metric are, in general, dependent on coordinates. Also note that this decomposition is in no way unique.

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The general answer is "yes". Gravitational waves are distortions of spacetime--changes in spacetime curvature--that propagate through spacetime as waves. Or, if that is still confusing, a spacetime with gravitational waves in it is just a spacetime with curvature that fluctuates in a wavelike manner--i.e., a 4-dimensional object with a geometry that fluctuates in a wavelike manner.

As far as i under stand it gravity is a property of space time, so gravity waves( being a table cloth would be dragged down a hole in the table) light is a wave that travels through space and not a property of space, or am i getting this completely wrong ?

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As far as i under stand it gravity is a property of space time, so gravity waves( being a table cloth would be dragged down a hole in the table)

Spacetime doesn't "move" or get "dragged". It's just a 4-dimensional geometric object. If gravitational waves are present, the geometry of this object has wavelike fluctuations in it. It's perfectly possible to have such a geometry where the wavelike fluctuations are everywhere, even if there are gravitating bodies like stars or black holes present.

It is still an open question if gravity has gravity. The stress energy tensor is unclear on that point. I would say ... I have no idea.

I don't think that's a proper way of saying it. Its not an open question but only a matter of different viewpoints. My idea is that the non-linearity of field equations means "gravity has gravity". But some people would interpret that differently(I don't know how! but if they can't do such an interpretation, their view would be unacceptable) and state that because gravity doesn't appear as a source in the field equations, "gravity doesn't have gravity". But to me, this is only a philosophical issue.

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It is still an open question if gravity has gravity. The stress energy tensor is unclear on that point. I would say ...

No; Shyan is correct. The stress-energy tensor is perfectly clear: it doesn't include any contribution from gravity. In that sense, gravity does not gravitate. (There are various pseudo-tensors that can be constructed that include a contribution from gravity as well as other kinds of energy, but they are not, as the "pseudo" indicates, actual tensors.)

However, since the Einstein Field Equation is nonlinear, when you look at solutions, you find that the complete absence of a source (i.e., zero stress-energy tensor) does not necessarily mean the complete absence of gravity (i.e., spacetime does not have to be flat). So in that sense, gravity can gravitate.

The "open question", if there is one, is which of those two viewpoints one considers to be "right". To me, the correct answer is "both"; the problem is expecting the question "does gravity gravitate?" to have a unique answer. It doesn't.

(Btw, thanks, Shyan, for linking to my much longer disquisition on this subject. )

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The usual definition uses linearized GR, so that the metric is written as ##g_{ab} = \eta_{ab} + h_{ab}##, and then writes the vacuum EFE in terms of this metric, throwing away terms quadratic or higher in ##h_{ab}##, and shows that ##h_{ab}## satisfies a wave equation. (You also have to choose an appropriate gauge for ##h_{ab}## .) You can also linearize about some other "background" metric ##\tilde{g}_{ab}##; this is what Shyan was describing in post #2. This works well as long as ##\tilde{g}_{ab}## is stationary, or at least varies on a much longer time scale than ##h_{ab}##.

In principle, since the EFE is nonlinear, one should also be able to look at higher orders in ##h_{ab}## to obtain self-interaction effects between the waves. However, no one expects this to have any practical utility any time soon; it's going to be tough enough just to detect the linear waves, let alone to ever detect higher-order effects.

The usual definition uses linearized GR, so that the metric is written as ##g_{ab} = \eta_{ab} + h_{ab}##, and then writes the vacuum EFE in terms of this metric, throwing away terms quadratic or higher in ##h_{ab}##, and shows that ##h_{ab}## satisfies a wave equation.

I suppose this is what confuses me. Are you saying that the definition is: if a space-time (vacuum) have a metric that (locally) can be written in the way above and the ##h_{ab}## satisfies a wave equation, then we say that there are gravitational waves in the space-time? It seems that your phrasing says that this holds for every spacetime, does that mean that it includes the trivial solution i.e. no waves.

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Are you saying that the definition is: if a space-time (vacuum) have a metric that (locally) can be written in the way above and the ##h_{ab}## satisfies a wave equation, then we say that there are gravitational waves in the space-time?

I'm saying that when you see physicists talk about gravitational waves, that's what they're talking about. But of course the description I gave is an approximation. There is no exact dividing line between what kinds of spacetime curvature are "gravitational waves" and what kinds are not. There are obvious extreme cases: the curvature of Schwarzschild spacetime is obviously not a "gravitational wave" since it is static; and the curvature described by the simplest plane wave solution to the wave equation for ##h_{ab}## is obviously the canonical example of a gravitational wave. But there are lots of possible in-between cases where the distinction is not so clear. Those in-between cases don't really have any practical significance, at least not now, so physicists don't spend much time on them.

Note that the background metric doesn't necessarily have to be vacuum or flat, although the case where it is both is the one that is most frequently discussed, both because it's the easiest to analyze and because any practical possibility of detecting GWs in the reasonably near future will be detecting waves in vacuum where the background spacetime is flat or close enough to it.

It seems that your phrasing says that this holds for every spacetime, does that mean that it includes the trivial solution i.e. no waves.

As a matter of logic, yes, it would--i.e., the case ##h_{ab} = 0## is, logically speaking, included. But physicists don't bother using the term "gravitational waves" to refer to the case where there are no waves.